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Unexpected result with right shift after bitwise negation
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I expected that below code will output 10 because (~port) equal to 10100101
So, when we right shift it by 4 we get 00001010 which is 10.
But the output is 250! Why?
int main()
{
uint8_t port = 0x5a;
uint8_t result_8 = (~port) >> 4;
//result_8 = result_8 >> 4;
printf("%i", result_8);
return 0;
}
c bit-manipulation
edited 16 mins ago
John Kugelman
249k54407460
asked 32 mins ago
IslamIslam
545
add a comment |
I expected that below code will output 10 because (~port) equal to 10100101
So, when we right shift it by 4 we get 00001010 which is 10.
But the output is 250! Why?
int main()
{
uint8_t port = 0x5a;
uint8_t result_8 = (~port) >> 4;
//result_8 = result_8 >> 4;
printf("%i", result_8);
return 0;
}
c bit-manipulation
edited 16 mins ago
John Kugelman
249k54407460
asked 32 mins ago
IslamIslam
545
add a comment |
I expected that below code will output 10 because (~port) equal to 10100101
So, when we right shift it by 4 we get 00001010 which is 10.
But the output is 250! Why?
int main()
{
uint8_t port = 0x5a;
uint8_t result_8 = (~port) >> 4;
//result_8 = result_8 >> 4;
printf("%i", result_8);
return 0;
}
c bit-manipulation
edited 16 mins ago
John Kugelman
249k54407460
asked 32 mins ago
IslamIslam
545
I expected that below code will output 10 because (~port) equal to 10100101
So, when we right shift it by 4 we get 00001010 which is 10.
But the output is 250! Why?
int main()
{
uint8_t port = 0x5a;
uint8_t result_8 = (~port) >> 4;
//result_8 = result_8 >> 4;
printf("%i", result_8);
return 0;
}
c bit-manipulation
c bit-manipulation
edited 16 mins ago
John Kugelman
249k54407460
asked 32 mins ago
IslamIslam
545
edited 16 mins ago
John Kugelman
249k54407460
asked 32 mins ago
IslamIslam
545
edited 16 mins ago
John Kugelman
249k54407460
edited 16 mins ago
John Kugelman
249k54407460
edited 16 mins ago
John Kugelman
249k54407460
249k54407460
asked 32 mins ago
IslamIslam
545
asked 32 mins ago
IslamIslam
545
asked 32 mins ago
IslamIslam
545
545
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
C promotes uint8_t to int before doing operations on it. So:
portis promoted to signed integer0x0000005a.
~inverts it giving0xffffffa5.- An arithmetic shift returns
0xfffffffa. - It's truncated back into a
uint8_tgiving0xfa == 250.
To fix that, either truncate the temporary result:
uint8_t result_8 = (uint8_t)(~port) >> 4;
or mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
answered 29 mins ago
ybungalobillybungalobill
46.1k1395162
you're right but i think C doesn't promote onlyuint8_tbut alsounsigned charbecause i tested it withunsigned chartoo and got the same result! Am i right?
– Islam
14 mins ago
2
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
11 mins ago
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
C promotes uint8_t to int before doing operations on it. So:
portis promoted to signed integer0x0000005a.
~inverts it giving0xffffffa5.- An arithmetic shift returns
0xfffffffa. - It's truncated back into a
uint8_tgiving0xfa == 250.
To fix that, either truncate the temporary result:
uint8_t result_8 = (uint8_t)(~port) >> 4;
or mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
answered 29 mins ago
ybungalobillybungalobill
46.1k1395162
you're right but i think C doesn't promote onlyuint8_tbut alsounsigned charbecause i tested it withunsigned chartoo and got the same result! Am i right?
– Islam
14 mins ago
2
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
11 mins ago
add a comment |
C promotes uint8_t to int before doing operations on it. So:
portis promoted to signed integer0x0000005a.
~inverts it giving0xffffffa5.- An arithmetic shift returns
0xfffffffa. - It's truncated back into a
uint8_tgiving0xfa == 250.
To fix that, either truncate the temporary result:
uint8_t result_8 = (uint8_t)(~port) >> 4;
or mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
answered 29 mins ago
ybungalobillybungalobill
46.1k1395162
you're right but i think C doesn't promote onlyuint8_tbut alsounsigned charbecause i tested it withunsigned chartoo and got the same result! Am i right?
– Islam
14 mins ago
2
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
11 mins ago
add a comment |
C promotes uint8_t to int before doing operations on it. So:
portis promoted to signed integer0x0000005a.
~inverts it giving0xffffffa5.- An arithmetic shift returns
0xfffffffa. - It's truncated back into a
uint8_tgiving0xfa == 250.
To fix that, either truncate the temporary result:
uint8_t result_8 = (uint8_t)(~port) >> 4;
or mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
answered 29 mins ago
ybungalobillybungalobill
46.1k1395162
C promotes uint8_t to int before doing operations on it. So:
portis promoted to signed integer0x0000005a.
~inverts it giving0xffffffa5.- An arithmetic shift returns
0xfffffffa. - It's truncated back into a
uint8_tgiving0xfa == 250.
To fix that, either truncate the temporary result:
uint8_t result_8 = (uint8_t)(~port) >> 4;
or mask it:
uint8_t result_8 = (~port & 0xff) >> 4;
answered 29 mins ago
ybungalobillybungalobill
46.1k1395162
answered 29 mins ago
ybungalobillybungalobill
46.1k1395162
answered 29 mins ago
ybungalobillybungalobill
46.1k1395162
answered 29 mins ago
ybungalobillybungalobill
46.1k1395162
46.1k1395162
you're right but i think C doesn't promote onlyuint8_tbut alsounsigned charbecause i tested it withunsigned chartoo and got the same result! Am i right?
– Islam
14 mins ago
2
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
11 mins ago
add a comment |
you're right but i think C doesn't promote onlyuint8_tbut alsounsigned charbecause i tested it withunsigned chartoo and got the same result! Am i right?
– Islam
14 mins ago
2
uint8_tis, very likely, a synonym ofunsigned charon your system. The promotion rules apply to all integral types smaller thanint.
– ybungalobill
11 mins ago
you're right but i think C doesn't promote only
uint8_t but also unsigned char because i tested it with unsigned char too and got the same result! Am i right?– Islam
14 mins ago
you're right but i think C doesn't promote only
uint8_t but also unsigned char because i tested it with unsigned char too and got the same result! Am i right?– Islam
14 mins ago
2
2
uint8_t is, very likely, a synonym of unsigned char on your system. The promotion rules apply to all integral types smaller than int.– ybungalobill
11 mins ago
uint8_t is, very likely, a synonym of unsigned char on your system. The promotion rules apply to all integral types smaller than int.– ybungalobill
11 mins ago
add a comment |
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